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Side-by-side: Late Cut-off.

Next the L. P. piston moves to cut-off at P3, but the cor-
responding movement of the H. P. piston is so small that it may
be neglected. Expanding regularly in all the vessels, we have :

il + -45 +

P       ---     n T  A      V     __________5_______I~_____

+ '3

=     28-6

s               .ij + '45 + if + "3 + ('I x 3'5)

The rest of the low-pressure diagram up to PL will be understood.
Following the H. P. diagram, P8 is compressed regularly to P4 so

" + '45

P4 = 28-6 x


. + * + *45

And now there is a sudden communication with the clearance
Lc, having a residual pressure of 18; therefore,

p . = 4*7 x (i + \ + '45) + '3 * l8 =    6
5                | + + -45 + -3

Then there is a gradual change of pressure, all three vessels
being in communication, but the curve is not a hyperbola, because
not only are the cylinders of different area, but the piston speed
varies considerably. At centre B strike the larger semi-circle, and
at centre A strike the smaller semi-circle, to represent respectively
the L. P. and H. P. cranks. Assume the H. P. compression
point 6 and join 6 A, then draw B 6 Nat right angles to 6 A ; also
divide the portions between 5 and 6 on each crank circle into
equal parts, and letter as shewn. Now the total volume at any
point between 5 and 6 can be found, it always being (Hc -h Lc -f R)
+ vol. in H. P. + vol. in L. P. Thus at P5, volume = (J+ -3 + -45)
+ 5+ 0=1'375. For any other position, w for example, the
volumes in H. P. and L. P. may be found by taking off both the
distances * * with dividers and measuring these by the scale FG,
We have not space to consider every point, but at P6 vol. will
clearly be "875 + *o8 + *7$75 = 1742. Then,

Intermediate points between P5 and P6 on H. P. diagram
being obtained, an arched curve is found as drawn. The H. P.
diagram is next completed by drawing a hyperbola through PG.
The L. P. curve from P5 to P6 must next be drawn. Now the